Question

The wavelength of $K_{\alpha }$ line for an element of atomic number 43 is ${}^{\prime }{\lambda }^{\prime }$ . Then the wavelength of $K_{\alpha }$ line for an element of atomic number 29 is

• A. $\frac{9}{4}\lambda$
• B. $\frac{43}{29}\lambda$
• C. $\frac{4}{9}\lambda$
• D. $\frac{42}{28}\lambda$

Answer 1

The correct option is A $\frac{9}{4}\lambda$

$\frac{1}{\lambda }=R(z-b{)}^2(\frac{1}{1}-\frac{1}{2^2})$

$\frac{1}{\lambda }=R(43-{)}^2\frac{3}{4}$

$\frac{1}{\lambda }=1323R$

$Now\frac{1}{\lambda }=R(29-1{)}^2\frac{3}{4}$

$\frac{\lambda }{{\lambda }^{\prime }}=588R$

$\frac{{\lambda }^{\prime }}{\lambda }=\frac{1323}{588}$

$\frac{{\lambda }^{\prime }}{\lambda }=\frac{9}{4}$